Multiple choice

A special dice is so constructed that the probabilities of throwing $1,2,3,4,5$ and $6$ are $\dfrac{1-k}{6},\dfrac{1+2k}{6},\dfrac{1-k}{6},\dfrac{1+k}{6},\dfrac{1-2k}{6}$ and $\dfrac{1+k}{6}$ respectively. If two such dice are thrown and the probability of getting a sum equal to $9$ lies between $\dfrac19$ and $\dfrac29$, find the set of integral values of $k$.

  1. Set of integral values of $k={0}$
  2. Set of integral values of $k={1}$
  3. Set of integral values of $k={0,1}$
  4. Set of integral values of $k=\phi$
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A Correct answer
Explanation

The sum of probabilities must be 1. Sum = (1-k + 1+2k + 1-k + 1+k + 1-2k + 1+k)/6 = 6/6 = 1. The condition 1/9 < P(sum=9) < 2/9 must be satisfied. Calculating P(sum=9) for different k values shows that only k=0 satisfies the condition.

AI explanation

The valid combinations for a sum of 9 on two dice are (3,6), (4,5), (5,4) and (6,3). Using the given probability distribution, the probability of this sum is 2 multiplied by [(1-k)/6 multiplied by (1+k)/6] plus 2 multiplied by [(1+2k)/6 multiplied by (1-2k)/6], which simplifies to (5 - 3k^2) / 18. For the individual probabilities to be valid, k must lie strictly between -0.5 and 0.5, making the only integer in this range and satisfying the given condition 0. The set of integral values is 0.

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